ios - 快速创建参数化字符串或宏？

Question

我是 Swift 的新手，我想知道如何在 Swift 中创建参数化字符串或宏？

就像我们在 Objective-C 中所做的那样:

#define SEARCH(name, limits) [[NSString stringWithFormat:@"https://someURL.com/search?term=%@&limit=%ld", name, limits]

如何在 Swift 中制定上述语句？

Answer 1

你可以定义全局函数:

func search(name: String, _ limits: Int) -> String {
    return "https://someURL.com/search?term=\(name)&limit=\(limits)"
}

同时不要忘记对 name 进行 URL 编码:

func search(name: String, _ limits: Int) -> String {
    let urlEncodedName = name.stringByAddingPercentEncodingWithAllowedCharacters(NSCharacterSet.URLQueryAllowedCharacterSet()) ?? ""
    return "https://someURL.com/search?term=\(urlEncodedName)&limit=\(limits)"
}